Re: [xAct] Simplification of Odd Parity Curvature Squared Terms

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Leo Stein

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Aug 26, 2024, 3:31:36 PMAug 26
to Sukruti, xAct Tensor Computer Algebra
Dear Sukruti,

Using RiemannToChristoffel is probably pretty inefficient. I attach an example showing how to implement the Bianchi identities of Riemann in the special case where 3 (or more) indices are contracted with the Levi-Civita tensor. This is a pretty restricted scenario, but if you're just interested in expressions arising from contractions of the dual of Riemann, it might be useful.

It's not enough to see the linear dependence between the last two scalars, which as you mention is due to a dimensionally-dependent identity. The Invar package indeed does have these dual Riemann invariants. This is also demonstrated in the attached notebook. Notice that RiemannSimplify has rewritten those two scalars so it's now clear that they are linearly dependent.

Best
Leo

On Wed, Aug 21, 2024 at 5:23 AM Sukruti <bansal....@gmail.com> wrote:

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dual Riemann.nb

Sukruti

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Aug 27, 2024, 8:56:12 AMAug 27
to xAct Tensor Computer Algebra
Dear Leo,

Thanks a lot again for replying to my query and sharing your notebook!

The way you have defined the Bianchi identities is quite efficient and for my purposes it is sufficient as well because I need to apply the identities only on terms which have an accompanying Levi-Civita tensor in them.

The RiemannSimplify command is a huge help! The two linearly dependent terms which remained in your example after applying RiemannSimplify, can actually be shown to be related to each other using the symmetry/anti-symmetry properties of the Riemann tensor indices and the first Bianchi identity alone. I observed this yesterday after I managed to derive one of those terms from the other by manual calculation. So I think those terms are not necessarily related via dimensionally dependent identities.

I have a follow-up question. If I use the command RiemannSimplify on your initial list of the four curvature square terms, i.e. before applying the Bianchi identities to it, then the first term gets simplified to an unrecognisable output which is 1/2 (2, 3) (4, 5, 7) (6, 9, 8) (shown in attached notebook).

Unfortunately when I apply RiemannSimplify on odd-parity curvature cube terms (shown in the attached notebook) even after applying the Bianchi identities, some of the terms give unrecognisable outputs of the kind mentioned above.

Any idea what could be the reason for such unrecognisable outputs and how to fix it?

Best,
Sukruti
Odd Parity Curvature Terms.nb
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